Question about Ellipse

Here is a problem in coordinate geometry, in particular about the ellipse.

A point moves such that the sum of the squares of its distances from two intersecting straight lines is constant. Prove that its locus is an ellipse and find the eccentricity in terms of the angle between the straight lines.

My solution:

Without loss of generality we may assume the two straight lines to be [itex]y = 0[/itex] and [itex]y = mx[/itex] where [itex]m = \tan\phi[/itex] ([itex]\phi[/itex] is the angle between the lines). Their point of intersection is thus the origin O(0,0).

Let the point whose locus is to be found be [itex]P(\alpha,\beta)[/itex]. The constraint on P is then,

[tex]\beta^2 + \frac{(m\alpha - \beta)^2}{m^2+1} = k^2 [/tex]

where k is some constant ([itex]k\epsilonR[/itex])

This after some rearranging and replacing [itex](\alpha,\beta)[/itex] with with general coordinates [itex](x,y)[/itex] yields
[itex]m^2x^2 - 2mxy + y^2(m^2+2) - k^2(1+m^2) = 0[/itex]
which when compared with the general second degree equation,
[itex]Ax^2 + 2Hxy + By^2 + 2gx + 2fy + c = 0 [/itex]
does turn out to be an ellipse.

However it is not in the standard form, so finding its eccentricity is not as easy. Now I understand that by rotating the coordinate axes we can bring the equation into such a form by a suitable choice of the rotation angle which causes the cross term (H) to disappear. However, I want to know if there is some other way out to find the eccentricity (or more generally to do this problem).

I don't see any other way to extract the excentricity,other than knowing the semiaxes...You can do that simply applying the theory of conics and the set of linear transformations which bring the conic to a known form,in this case an ellipse.

This is squashing the coordinates, and so will transform a circle into an ellipse

Yes but the problem is to extract the eccentricity from the equation without reducing it to a standard form via rotation. However, as the problem stands now, I do not think there is any other method than to do it the brute force way. Thanks for your help though.